Q72P Ball B, moving in the positive d... [FREE SOLUTION]

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Chapter 9: Q72P (page 253)

Ball B, moving in the positive direction of an xaxis at speed v, collides with stationary ball Aat the origin. Aand Bhave different masses. After the collision, Bmoves in the negative direction of the yaxis at speed v/2 . (a) In what direction does Amove? (b) Show that the speed of A cannot be determined from the given information.

Short Answer

Expert verified

a) Ball A can move at angle $蠒 = 27^{掳}$ .

b) Proved

Step by step solution

Step 1: Given Data

$v_{B} = v$

$v_{B}^{'} = v / 2$

Determining the concept

Let, ball B be moving in a positive direction with speed v and ball A be at the origin. But after collision, the direction of ball B will be in negative y鈥� axis with half speed. At the same time, the speed of ball A may reduce to half. As ball A is stationary and both the balls contain different masses, logically, ball A will move inthedirection along x-positive axis. Here, linear momentum is conserved.According to conservation of linear momentum, momentum that characterizes motion never changes in an isolated collection of objects.

Formula is as follow:

Initial momentum = Final momentum

(a) Determining the direction in which A can move

Conservation of linear momentum along x and y axis can be given as,

$m_{B} v_{B} = m_{B} v_{B}^{'} \cos 胃 + m_{A} v_{A}^{'} \cos 蠒 . . . . . . (1) 0 = m_{B} v_{B}^{'} \sin 胃 + m_{A} v_{A}^{'} \sin 蠒 . . . . . . (2)$

where, $m_{B}$ denotes mass of ballB

$v_{B}$ denotes velocity of ball B,

$V_{b}^{'}$ denotes velocity of ball B after collision,

$m_{A}$ denotes mass of ball A,

$v_{A}$ denotes velocity of ball A,

$v_{A}^{'}$ denotes velocity of ball A after collision.

Now, to find the direction of ball A after collision, only consider A part from equation (2). Therefore,

$0 = m_{B} v_{B}^{'} \sin 胃 + m_{A} v_{A}^{'} \sin 蠒 m_{A} v_{A}^{'} \sin 蠒 = m_{B} \frac{v}{2} . . . . . (3)$

Also, for x- direction, equation (1) will be,

$m_{B} v_{B} = m_{B} v_{B}^{'} \cos 胃 + m_{A} v_{A}^{'} \cos 蠒 m_{A} v_{A}^{'} \cos 蠒 = m_{B} v_{B} . . . . . (4)$

Dividing equations (3) and (4), therefore,

$\tan 蠒 = 1 / 2 蠒 = 27 掳$

Thus, ball A moves at an angle of $27 掳$ .

(b) Showing that speed of A cannot be determined from the given information

To find the velocity of ball Aafter collision, from equation (1) or (2), values for masses of ball A and B are needed. Since, these values are not given, the speed of ball A cannot bedetermined.

Hence,it is shown that speed of A cannot be determined fromthegiven information.

Therefore, using the law of conservation of linear momentum, the direction of motion of the ball after the collision can be determined.